summarization heat transfer.docx
TRANSCRIPT
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(4 marks)
Solution Q1 (b)
AnalysisNoting that the cross-sectional area of the spoon is constant and measuring x from
the free surface of water, the variation of temperature along the spoon can be expressed as
T x T
T T
a L x
aLb
( ) cosh ( )
cosh
where
2
m000026.0)m100/2.0m)(100/3.1(
m03.0)m100/2.0m100/3.1(2
cA
p
1-
2
2
m162.36)m000026.0)(CW/m.15(
)m03.0)(C.W/m17(
ckA
hpa
Noting that x =L = 18/100 = 0.18m at the tip and substituting, the tip
temperature of the spoon is determined to be
C25.2=5.335
1)25(95+C25=)18.0162.36cosh(
0cosh)25(95+C25=
cosh
)(cosh)()(
aL
LLaTTTLT b
Therefore, the temperature difference across the exposed section of the spoon handle is
C69.8 C)2.2595(tipTTT b
Tb
L = 18 cm
(2 marks)
(2 marks)
(1 marks)
(1 marks)
(4 marks)
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Q1(B)
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Solution Q2 (a)
The combination of criteria for a fin gives rise to a dimensionless group ht/k, termed thefin
Biot number, Bi.
For a good fin, Bi no. has to be small.
Heat has to flow through half the fin thickness in the transverse direction. Thus t in the fin
Biot number is half the fin thickness. For cylindrical fins of diameter 2r, t = r.
(4 marks)
Solution Q2 (b)
PropertiesThe thermal conductivity, density, and specific heat
of the milk at 20C are k = 0.607 W/m.C, = 998 kg/m3, and
Cp= 4.182 kJ/kg.C (Table A-9).
AnalysisThe characteristic length and Biot number for the glass of milk are
1.0>076.2)CW/m.607.0(
)m0105.0)(C.W/m120(
m01050.0m)03.0(2+m)m)(0.0703.0(2
m)07.0(m)03.0(
22
2
2
2
2
2
k
hLBi
rLr
Lr
A
V
L
c
oo
o
sc
For the reason explained above we can use the lumped system analysis to determine how long
it will take for the milk to warm up to 38C:
min5.8s348
teeTT
TtT
LC
h
VC
hAb
tbt
i
cpp
s
)s002738.0(
1-
3
2
-1
603
6038)(
s002738.0m)C)(0.0105J/kg.4182)(kg/m(998
C.W/m120
(3 marks)
(3 marks)
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Therefore, it will take about 6 minutes to warm the milk from 3 to 38C.
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Q2(a)
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Q2(b)
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Solution Q3 (a)
(i)
Explain how does the convection heat transfer coefficient,hdiffer from the thermal
conductivity, kof a fluid.
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The value of the convection heat transfer coefficient depends on the fluid motion as well as the
fluid properties. Thermal conductivity is a fluid property, and its value does not depend on the
flow. (2 m)
(ii)
For the parallel flow over the flat plate, sketch the temperature profile for the surface
temperature larger than fluid temperature. (3 m)
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Q3(b)
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Solution Q4 (a)
Define natural convection and explain the situation of natural convection with an appropriate
illustration. (6 marks)
Natural convectionis a convection heat ransfer occurs due to temperature differences which affect
the density, and thus relative buoyancy, of the fluid. (2 m)
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or any else which suitable (1m)
- Heat from hot object (hot boiled egg) will heat up the colder air adjacent to the hot object
surface. (1 m)
- These air becomes lighter (less dense) and will rise up this leading to bulk fluid movement.
(2m)
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Solution Q5 (a)
(i) The heat exchangers usually operate for long periodsof time with no change in their operating conditions,
and then they can be modeled as steady-flow devices.
(ii) The mass flow rate of each fluid remains constant and the fluid properties such as temperature and
velocity at any inlet and outlet remain constant.
(iii) The kinetic and potential energy changes are negligible. The specific heat of a fluid can be treated as
constant in a specified temperature range.
(iv) Axial heat conduction along the tube is negligible.
(v) The outer surfaceof the heat exchanger is assumed to be perfectly insulatedso that there is no heat lossto
the surroundingmedium and any heat transfer thus occurs is between the two fluids only.
Solution Q5 (b)
Glycerin is heated by hot water in a 1-shell pass and 13-tube passes heat exchanger. The mass flow
rate of glycerin and the overall heat transfer coefficient of the heat exchanger are to be determined.
Properties The specific heats of water and glycerin are given to be 4.18 and 2.48 kJ/kg.C,
respectively. [ ( )] (5Q mC T T p in out water kg / s)(4.18 kJ / kg. C)(100 C C) = 940.5 kW55
The mass flow rate of the glycerin is determined from:
kg/s9.5
C]15CC)[(55kJ/kg.(2.48
kJ/s5.940
)(
)]([
glycerin
glycerin
inoutp
inoutp
TTC
Qm
TTCmQ
(2 marks)
The logarithmic mean temperature difference for counter-flow
arrangement and the correction factor F are
T T TT T T
h in c out
h out c in
1
2
100 5555 15
, ,
, ,
C C = 45 CC C = 40 C
Glyceri
n
55C
(2 marks)
(2 marks)
(1 marks)
(1 marks)
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T
T T
T Tlm CF,
ln( / ) ln( / ).
1 2
1 2
45 40
45 4042 5 C
77.0
89.0100555515
53.010015
10055
12
21
11
12
F
ttTTR
tT
ttP
The heat transfer surface area is
2m0.94=m)m)(2015.0(10 DLnAs
Then the overall heat transfer coefficient of the heat exchanger is determined to be
C.kW/m30.6 2
C)5.42)(77.0)(m94.0(
kW5.9402
,,
CFlmsCFlms
TFA
QUTFUAQ
(2 marks)
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(1 marks)
(1 marks)
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Q5(B)
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Solution Q6 (a)
The value of effectiveness increases slowly with a large values of NTU (usually larger than 3). Therefore,
doubling the size of the heat exchanger will not save much energy in this case since the increase in the
effectiveness will be very small.
Solution Q6 (b)
Cold water is heated by hot water in a heat exchanger. The net rate of heat transfer and the heat transfer
surface area of the heat exchanger are to be determined.
Properties The specific heats of the cold and hot water are given to be 4.18 and 4.19 kJ/kg.C, respectively.
Analysis The heat capacity rates of the hot and cold fluids are
C m C
C m C
h h ph
c c pc
,
(0.25 kg / s)(4180 J / kg. C) W/ C
(3 kg / s)(4190 J / kg. C) W/ C
1045
12 570
Therefore, CW/1045min cCC
and C C
C min
max ,.
1045
12 5700083
Then the maximum heat transfer rate becomes
( ) ,max min , ,Q C T T h in c in (1045 W/ C)(100 C -15 C) W88 825
The actual rate of heat transfer is
W31,350 )C15C45)(CW/1045()( ,, outhinhh TTCQ
Then the effectiveness of this heat exchanger becomes
Q
Qmax
,
,.
31350
88825035
The NTU of this heat exchanger is determined using the relation in Table 13-5 to be
438.01083.035.0
135.0ln
1083.0
1
1
1ln
1
1
CCNTU
Then the surface area of the heat exchanger is determined from
2m0.482
C.W/m950
)CW/1045)(438.0(2
min
min U
CNTUA
C
UANTU
Hot Water
Cold Water
15C
45C
(5 marks)
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Q6(a)
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Q6(b)